Question: $\dfrac{ -8s + 3t }{ 9 } = \dfrac{ 6s + 4u }{ 4 }$ Solve for $s$.
Solution: Multiply both sides by the left denominator. $\dfrac{ -8s + 3t }{ {9} } = \dfrac{ 6s + 4u }{ 4 }$ ${9} \cdot \dfrac{ -8s + 3t }{ {9} } = {9} \cdot \dfrac{ 6s + 4u }{ 4 }$ $-8s + 3t = {9} \cdot \dfrac { 6s + 4u }{ 4 }$ Multiply both sides by the right denominator. $-8s + 3t = 9 \cdot \dfrac{ 6s + 4u }{ {4} }$ ${4} \cdot \left( -8s + 3t \right) = {4} \cdot 9 \cdot \dfrac{ 6s + 4u }{ {4} }$ ${4} \cdot \left( -8s + 3t \right) = 9 \cdot \left( 6s + 4u \right)$ Distribute both sides ${4} \cdot \left( -8s + 3t \right) = {9} \cdot \left( 6s + 4u \right)$ $-{32}s + {12}t = {54}s + {36}u$ Combine $s$ terms on the left. $-{32s} + 12t = {54s} + 36u$ $-{86s} + 12t = 36u$ Move the $t$ term to the right. $-86s + {12t} = 36u$ $-86s = 36u - {12t}$ Isolate $s$ by dividing both sides by its coefficient. $-{86}s = 36u - 12t$ $s = \dfrac{ 36u - 12t }{ -{86} }$ All of these terms are divisible by $2$ Divide by the common factor and swap signs so the denominator isn't negative. $s = \dfrac{ -{18}u + {6}t }{ {43} }$